Lec 14 | MIT 3.091SC Introduction to Solid State Chemistry, Fall 2010

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PROFESSOR: OK, so this takes us into the paper by Heitler
and London, where, by using the concept of linear
combination of atomic orbitals into molecular orbitals for a
large number of atoms, instead of splitting one plus one
atoms, making a bond into two energy levels of bonding and
AN antibonding, we take a large number of atoms, bring
them together, and we get a large number of bonding
orbitals and a large number of antibonding orbitals.
And they have to lie within a zone of about 5 to 10 electron
volts, which means if you put Avogadro's number of these
things together, you're going to end up with an energy
difference between successive energy levels on the order of
about 10 to the minus 20 electron volts, or 10 to the
minus 40 joules.
And for all intents and purposes, that is a continuum,
and hence, the term band.
And now we start populating this with electrons, and each
one of these is a separate state.
And electrons go in two by two, just like Noah's Ark, all
the way up-- either halfway up or all the way up.
But now the energy difference is very, very tiny.
And a second instance, we looked at insulators where--
for example, in diamond--
where the bonds are really strong.
So that means if the atomic orbital is roughly at the
middle here, the depression to make the bonding orbital is
very, very great, and very negative.
And therefore, the elevation to the antibonding orbital is
great and relatively positive.
And when we start putting Avogadro's number of these
things together, we still end up with the 10 to
the minus 40 joules.
But when we get to the top of the bonding orbitals, we have
still got 3 or 5 or 5 electron volts to get to the bottom of
the antibonding orbitals, hence the formation of the
band gap with an energy of--
as I mentioned.
So up here, we have the conduction band where
electrons are free to move about.
And this is derived from the antibonding orbitals.
And down here, we have the valence band, which consists
of the bonding orbitals derived from sigma and pi.
And then the semiconductor is sort of like an insulator--
as they say in California, it's like an insulator-- but
the band gap isn't quite so severe.
It's down on the order of 1 to 3 electron volts, and that
gives it unique properties.
What's magical about 1 to 3 electron volts?
Number one, visible light.
Visible light is 2 to 3 electron volts.
So visible light shines right through an insulator, because
visible light has an energy of about 2 to 3 electron volts.
If that's the energy of 2 to 3 electron volts, and this is 6,
it zooms right through.
But if it's 2 electron volts and this is 1 electron volt,
we're going to get thermal excitation.
I know Professor Ballinger talked about thermal
excitation the last day and what happens, and it's very
reminiscent of excitation in the case of the Bohr model.
Photoexcitation across the band gap.

Incident photon with energy greater than the
energy of the band gap.
And it's the same drill as you've seen before.
That's why those early lessons on--
photon comes in or electron comes in, strikes the Bohr
atom, electron rises up, momentarily sits there, falls
back down, photon emitted.
It's the same set of ideas.
Only instead of moving from n equals 1 to n equals 2 in the
Bohr atom, you're moving from the valence band to the
conduction band, sitting in the conduction band
momentarily, you fall back down, and you get a
So that was covered last day.
And I know towards the end of the lesson, he started talking
about thermal excitation.
So let's look at that in a little more detail.
Thermal excitation.
So thermal excitation of what?
Thermal excitation of electrons across the band gap.
So this is matter energy interaction, only in this
case, it's thermally inspired.
So let's draw our cartoon again.
Up here, we have the conduction band, and below we
have the valence band.
So in the conduction band we have free electrons.
In the valence band is where all the bonds live.
And we've got an energy gap here of some value.
Could be semiconductor, could be insulator.
In this case, we're going to use visible light.
And we've gone through the analysis with incident photon
and the emission.
In this case, I want to look at thermal excitation.
So with thermal excitation, I start with an electron down
here in a bond.
This electron lives in a bond.
And with thermal excitation, the electron jumps out of the
bond and up into the conduction band.
So I'm going to designate the electron in the conduction
band with an e minus and a circle around it.
And it leaves behind a broken bond.
As you know, every bond has two electrons in it.
Well, one of the electrons left the bond, leaving a hole.
Little h with a plus sign.
This is a broken bond, and a broken bond is highly unstable
and very mobile.
If I put a potential across here.
If I put this material-- and I'm doing a mixed metaphor--
this is an energy level diagram.
But now I'm going to put a couple of plates across an
energy level diagram.
But you're smart.
You can understand this.
So this is not energy.
Now I'm going to make the right electrode negative and
I'm going to make the left electrode positive.
What would happen?
Well, the electrons in the conduction band are going to
be attracted to the positive electrode or repelled by the
negative electrode, but the complementary situation occurs
in the valence band.
This broken bond is like a hot potato, and it has an
effective charge.
This is a zero in a land of minus one.
A broken bond is a zero in a land of minus one.
So therefore, if I make the right side negative, this hole
is going to be drawn.
So I have two forms of electronic conduction.
I have electrons moving to the left in the conduction band
and I have holes moving to the right in the valence band.
So for every thermal excitation I get two carriers.
Every excitation of electron from valence band to
conduction band generates two carriers.
Carriers of what?
Carriers of electrical charge.
Two carriers of electrical charge.
And what are they?
The electron in the conduction band and the hole--
little h with a plus sign--
in the valence band.
So that's how thermal excitation works.
And you can see that the numbers have to be equal.
So we write, therefore, that the populations are equal.
The number of electrons in the conduction band equals the
number of holes in the valence band.
Now electrical engineers have a similar expression, but they
don't care about numbers and electrons and holes, they care
about the functionality.
And the electron is a carrier of negative charge and the
hole is a carrier of positive charge.
So the electrical engineering books will write, thermal
excitation gives rise to pair generation.
Pair of what?
Pair of charge carriers.
And the population of negative charge carriers equals the
population of positive charge carriers, which in chemistry
is the number of electrons equals the number of holes, or
the number of broken bonds.
So that's the way it works, but there's a conundrum here.
Thermal excitation is based upon the thermal energy, and
we know that the average thermal energy in an
environment is roughly given--
I mean there are fancier calculations--
but to a first order, I can approximate it as the product
of the Boltzmann constant and absolute temperature.

So at room temperature--
room temperature, I like to say, about 300 Kelvin.
300 Kelvin is 27 Celsius.
That's kind of a warm room, but 300 is a nice number.
All right?
So if you take 300 Kelvin, you get on the order of about 1/40
of an electron volt.
1/40 of an electron volt, which is a lot less than one
electron volt, which is roughly the band gap of a
So how is it that we get any thermal excitation of data
indicate that there is some thermal excitation even at
room temperature?
So the question is, if the average thermal energy is a
fortieth of electron volt according to this-- and this
is correct--
and the band gap is one eV, then it should be no thermal
But we know what happens.
So, one eV, which I'm going to say is Eg.
All right?
So how do we get any thermal excitation under these
So there's an answer to this.
And for this we go to merry old England and James Clerk
Maxwell, that you know or will come to know in 8.02 as the
annunciator of Maxwell's equations of
electricity and magnetism.
He also thought about physical chemistry, and in particular
he thought about the physical chemistry of gases.
And what did he teach us about gases?
He said, if you look here in this room, you see nitrogen
and oxygen and all the other constituents of air.
But to a first approximation, we have four parts nitrogen,
one part oxygen.
The room is roughly 300 Kelvin.
What Maxwell taught us is that velocities of the gas species
zooming around in this room are not all equal.
Some of the gas species are moving around as though
they're at 500 Kelvin.
And some of them are moving around very slowly as though
they're at 100 Kelvin.
On average, it's 300 Kelvin, because we can
feel it as 300 Kelvin.
But some of them are zooming around.
Some are really lazy and they're just moving along
really slowly.
And furthermore, he was quantitative about it.
He said, I'm going to tell you what the distribution of
energies is.
So let's get that on the board for starters.
That in any ensemble of gas species, could be gas atoms or
it could be gas molecules or gas compounds, in any ensemble
of gas species, that all molecules--
because you can always say the atom is a degenerate form--
all molecules do not have the same velocity.

And furthermore, he gave us the distribution.
He calculated the distribution of velocities.
So that was about 1859.
Forgive me here.

And then Boltzmann came along in 1872.
And Boltzmann generalized this idea.
He made the link between gas velocity and temperature.
So, he took the idea and he said, if the gas molecules
move at different velocities, that means they
have different energies.
And why do we call this the Boltzmann constant?
Because Boltzmann made the link between particle energy
and temperature.
So he said there's a variation of instant temperature.
All right?
So Boltzmann generalized Maxwell's idea to energy and
then ultimately temperature.
And so I'm going to show you, instead of the Maxwell
distribution, I'm going to give you the Maxwell-Boltzmann
But before I do, I want to make sure you understand the
math and appreciate what I'm going to plot for you.
So we just had a test, right?
In fact, we can go back to the slide.
All right.
So what do we have here?
I could plot on this axis, the abscissa, the grade--
and on the ordinate, I'll plot the number of people who have
any given grade--
n of g.
Now the class av was 66, all right?
This is going to go from 0 to 100.
Now what we see here is-- and by the way, I'm going to
normalize it.
Instead of dealing with 463 or whatever, I'm going to divide
by the total class.
So the number is going to go from 0 to 1.
So this a normalized distribution.
What I'm teaching you applies all through science.
As I told you, this is the most important class.
Not because I'm teaching, but because the subject matter is
general, just as Boltzmann did.
So let's look at the normalized distribution.
Now the class av was 66.
Now we didn't have everybody getting 66.
That was not the distribution.
That's one solution to the problem, isn't it?
Instead, what we got was we got that.
Now one possibility would be this one,
the bell-shaped curve.
Goes like this.
That's bell-shaped.
This is Gaussian.
This is a Gaussian distribution, and its
functionality is y equals e to the minus x minus x average,
quantity squared.
Now, to have a normal distribution, you have to have
a class of normal people, so that didn't happen on this
test. That's a joke.
God, you're so serious.
Have you no fun in your lives?
Have you ever laughed?
PROFESSOR: Overachievers, you.
So we're not going to have a Gaussian distribution.
We're going to have that one there.
It's kind of a skewed Gaussian with a long tail.
All right, so that's the one thing.
But what Boltzmann gave us, instead of the number of atoms
of a given grade, he plotted it as a function of energy.
So instead, the Maxwell-Boltzmann distribution
has energy as the ordinate instead of the grade, and up
here, the number of atoms--
or more importantly--
the atom fraction.
The fraction of atoms in the distribution
that have that energy.
And it doesn't look like a normal distribution.
It looks sort of off-normal.
Skewed, like this.
It rises to a peak and it has a long tail that
moves off to the right.
So this is the Maxwell-Boltzmann
It's not y equals x minus x bar squared.
So this is Maxwell-Boltzmann distribution of energies.
And furthermore, what we see here is that the area under
this line has to equal 1.
If you integrate the fraction from 0 to infinity, because
this asymptotically goes out to infinity, the integral
under this line is 1.
Area equals 1, which is the integral of ndE.
Now, because of the asymmetry here, clearly this is the
average energy.
The average energy isn't the maximum energy.
Whereas in the Gaussian--
let's keep the Gaussian over here, just for grins and
chuckles-- so the Gaussian looks like this.
It's symmetric.
And this is the n of g max and also a g average, whereas
here, the average is a little bit to the right.
A little bit higher.
So now, if this is the distribution, and this at room
temperature is 1/40 of an electron volt, and way, way up
here is the band gap energy, there's a tiny but non-zero
fraction of this distribution that has energy in excess of
the band gap energy.
And so this tiny fraction of the total distribution has the
thermal energy to allow this excitation and pair
generation to occur.
And furthermore, this is a function of temperature.
This is a function of temperature.
So E average is related to temperature.
So, I'll say E average 1 at T1.
And then what I'm going to do is I'm going to heat this to a
higher temperature, so I'll use red chalk to
indicate it's hotter.
And what happens to the distribution under an increase
in temperature?
Well, the average had better move to the right, and that's
what happens.
But it doesn't just move to the right, it skews a little
bit and broadens.
So in point of fact, at a higher temperature,
it looks like this.

So this is now E average 2 at T2 where T2 is
greater than T1.
T2 is greater than T1.
OK, so that's the Maxwell-Boltzmann
But here's where the interesting stuff occurs.
So I'm going to blow up this high energy end of the
So again, this is going to be n of E
normalized, and this is E.
And what we're going to find here is that at T1--
this is the curve at T1, and the area under the line is
shown here--
and then--
not to scale--
we're going to put a break in this.
And I'll show you how much of a break is needed.
And then, at T2 I'm going to have this come down.
This is T2.
And the area under the line T2 captures the
T1 plus all of this.

And so what we see, is that for a modest increase in
temperature, this maximum shifts modestly, but the area
under the line in excess of this critical
energy increases radically.
And that's the powerful leveraging factor of increase
in temperature.
And I've done some calculations just to show you
what this does.
The functionality of the Maxwell-Boltzmann distribution
is e to the minus some critical energy divided by the
product of Boltzmann constant temperature.
So we can put in a value here-- eg--
what have you.
All right.
So I chose for silicon--
for silicon, I know the band gap energy is 1.1 eV--
and I have it at room temperature--
when I go through the calculation--
the fraction underneath that line is 10 to the minus 19.
One part in 10 to the minus 19.
This is the area.
Area under the curve is greater than the band gap
energy of 1.1 eV.
So then I said, let's go up to the melting point.
I'm not going to melt the silicon, I'm just going to
bring it up to its melting point but not melt it.
So it's like ice cubes at 0 Celsius.
It's still solid.
The melting point of silicon, you can look in your Periodic
Table, it's over 1400 degrees C.
And this is 10 to the minus 4.
10 to the minus 4.
Area under curve: da da da da.
So that means that the red area--
A2, the red.
Actually, let's use red chalk so that everybody is in tune.
A2 to A1 is equal to 10 to the 15, whereas T2 over T1--
well, red chalk.
It's my chance to use red chalk and I'm not going to
squander it.
T2 divided by T1--
it's about 1700 Kelvin divided by 300 Kelvin--
is about 6.
So for a change in temperature by a factor of 6, I get a
change in population of 10 to the 15.
So you can see how modest increases in temperature have
a huge impact on thermal excitation.
So now you say, if we wanted to make a device, we can't
heat our computers up to 1600 degrees, 1500 degrees C in
order to get an adequate number of charge carriers.
There has to be another way to get charge carriers.
I'm going to show you a third way that's
called chemical promotion.
So now I want to talk about charge carrier generation.
Charge carrier generation by change of chemistry, and it's
going to be specifically by the introduction of
Charge carrier generation by introduction of impurities.
You might say, gee, aren't impurities bad for a system?
Well, I want to tell you, there are two kinds of
impurities in this world.
There are bad impurities which we call contaminants, and
that's what you're commonly taught to think about.
That's bad.
But there are good impurities, and these are called dopants.
Dopants are impurities that we are happy
to have in our system.
And so by the introduction of impurities of value, we can
lead to engineered materials.

So I'm giving you the introduction to high
technology, where we're going to control the engineering
properties of a material through control of chemistry.
So that's chemical composition tailored--
for specific--

for targeted values of properties.
And that's the essence of material science.
We control the chemistry.
We control the properties by controlling the chemistry.
So what I want to do is give a vivid example of this, and
it's not going to be an isolated example.
It's going to be the example that pertains to all of the
microelectronic devices in our world.
So that's silicon.
So let's look at how doping works in silicon.
It also applies to germanium, but we don't use much
germanium in devices.
It's dominantly silicon.
So what we're going to do is dope.
The gambit here is to dope--
that's introduce an impurity--
with aliovalent impurity.
What do I mean by aliovalent?
You know the word alias?
It's another name.
Alio in Latin means other.
So we're going to dope silicon with something that has a
different valence from that of silicon.
Can't before.
So let's look at the simplest example.
And I'm going to put phosphorus into silicon.
I'm going to dope phosphorus into silicon, and phosphorus
is Group 5, or more properly, IUPAC calls it Group 15.
So we're going to put Group 15 element into silicon, which is
Group 14 element.
So we call phosphorus a supervalent dopant, meaning it
has a valence higher than that of the silicon.
So it's a supervalent dopant.
Supervalent dopant.
So now let's look at how that supervalent dopant works.
I'm going to show you some silicon here, and I'm going to
give you the silicon.
This is sp3 hybridized.
it's sitting here as a single crystal.

Here are three silicons.
So it's a single crystal.
All these silicons on the bottom row line up.
All the silicons in the center line up.
And there are silicons at the top and they all line up.
So that if I look at it on the edge, it's a perfectly,
atomically smooth system.
And I have one of these.
I know Professor Ballinger made reference to it last day.
This is last-generation technology.
It's only eight-inch diameter.
But this is an eight-inch diameter silicon wafer.
So this started as a giant salami, eight inches in
diameter-- probably two meters long.
Single crystal.
Every atom here is adjacent to the next atom according to the
rules of atomic arrangement.
And when this thing is cut and polished,
it's flat to one atom.
It is atomically flat.
And it makes a fantastic shaving mirror.
Actually Dave, can you cut to this?
Yeah, here we go.
Here's the penny.
And this is silicon here.
It's going to drive this thing nuts because it can't focus on
this thing, because it's so--
there we are.
So this was a part of a giant salami, and now it's been cut.
And obviously you can't cut it one-atom thick.
It's some submillimeter thickness, because you need to
have some mechanical strength to it.
But the surface is absolutely flat to one atom.
All right.
And that's what we're going to dope into.
We're going to dope into this thing, and I'll show you how
we're going to dope it later.
Actually, why don't we cut to that.
Here's something that's already been doped and made
into devices.
So there's the plain old silicon, which you're having a
hard time seeing anything except a little bit of
And now here's the thing that's been processed, so
let's zoom.

So now you can see all the features.
So these features involved gas phase reaction to introduce
dopant atoms into the silicon lattice.
And then we're going to chop these little things up and put
them in your cell phones and in your laptops and whatever
other devices are out there.
So this is what we're talking about, and if we live long
enough, we'll see one of these.
This is a Pentium chip.
So where's the chip?
The chip is underneath this piece of gold.
Where did I put my pointer?

Ah, it's over here.
All right.
So the actual silicon device is underneath this.
And all of these are this spider's web of leads.
It's trying to access all the tiny devices.
It doesn't do you any good to have a device density that's
greater than your access density.
So the technology that goes into this is phenomenal.
Plus, these things all generate heat.
When you put all these tens of thousands of devices into
something the size of your fingernail, you've got a
toaster oven working here, and you have to get that heat out.
And this is several generations old.
What you have in your machines today is
even denser than this.
And supercomputers?
Forget it.
You have to have those things so aggressively cooled or else
they'll heat up and it'll be just like a
light bulb, just bursting.
So that's what's going on inside.
So let's see how that happens.
Let's cut back to-- well, you can leave that up.
It's a pretty image and will probably soothe the nerves of
the students.
So now I'm going to put phosphorus in here.
I'll put phosphorus in here, but where does phosphorus go?
Phosphorus goes onto a silicon site.
So it actually comes in and sits here.
Phosphorus comes and sits at a silicon site.
Now what do we know about the number of bonds that
phosphorus likes to form?
It has five valence electrons.
So it forms one, two, three, four bonds, and
it has a fifth electron.
And that fifth electron is sitting there somewhere.
It's sitting there somewhere in the lattice.
All right.
It's sitting somewhere in the lattice because it has no
place to go.
Now in energy space, I know where it goes.
So here's the conduction band of silicon, and this is the
valence band.
And this is all of the silicon host crystal, because the
dopant is in a tiny, tiny amount.
We're talking parts per million.
So for all intents and purposes, it's
still a silicon crystal.
It has tiny amounts of phosphorus in it.
So this is the band gap of silicon, where this is about
1.1 electron volts.
And where does this fifth electron go?
This fifth electron sits up in here in the conduction band,
doesn't it?
And every time I introduce a phosphorus, I have a fifth
electron that goes in the conduction band.
But I don't have to generate a hole.
I don't need to break bonds in order to put electrons in the
conduction band.
Because the fifth electron is like the fifth wheel.
You know?
It goes in the conduction band.
So there's no generation of holes here.
All right.
And furthermore, I can control the conductivity.
At the end of last lecture, Professor Ballinger showed you
that the conductivity of a substance is related to the
number of charge carriers.
So if I want to double the number of phosphoruses, I will
double the number of electrons in the conduction band,
independent of temperature.
All right.
So for every phosphorus dopant atom, we get one electron in
the conduction band.
We get one electron in the conduction band and no holes.

No holes in the valence band.
It's direct.
It's direct.
All right.
Now, electrical engineers would look at this and they'd
say, I don't care if it's phosphorus.
They don't care about the chemistry, they care about the
What kind of carriers am I throwing into this crystal?
For an electrical engineer, there are only
two kinds of carriers.
Positive carriers and negative carriers.
For every phosphorus, what's the outcome?
A negative carrier.
So this is now doped silicon.
It generated negative carriers, so
this is called n-type.
This is now n-type silicon.
And its properties are governed by the population of
these-- because I'm going to show you in a second that the
number of these far exceeds the number that are thermally
so the properties here are called extrinsic.
So now we have a crystal that is exhibiting it's extrinsic
behavior, thanks to the doping with the supervalent impurity.

One more thing.
One more thing.
Something very, very cool here.
You see this electron?
It's sitting here somewhere in the--
now this is Cartesian map.
I'm looking through a scanning electron microscope.
I see all the silicons.
I get the phosphorus.
And there's an electron in here somewhere.
And this has 15, right?
This has 15 protons.
This has 14 protons.
So in a land of 14 plus, 15 plus is relatively
positive, isn't it?
The phosphorus is locally positive in comparison with
the rest of the silicon lattice.
And it's pinned, because it's covalently bonded.
And there's an electron sitting here
with no place to go.
Do you know of a model that describes the behavior of a
one-electron system around a pinned positive nucleus?
AUDIENCE: The Bohr model.
PROFESSOR: Bohr model?
Just for grins and chuckles, what if we were to use the
Bohr model to describe the behavior of that electron?
You can say it's crazy.
That doesn't make any sense, because that's not a
one-electron system.
But what if we modeled it as though it were?
I'm going to-- just for grins and chuckles-- get a sense of
what that ground state value is.
And recall that in hydrogen, the ground state--
E1, ground state of the electron--
is equal to minus k.
Which is minus me to the fourth over 8 epsilon naught
squared times Planck constant squared.
Which is minus 13.6 electron volts
So what is the energy of that electron there if we use the
Bohr model?
Well, we have to modify it a little bit, because the
electron isn't moving in a vacuum.
Depending on how far this electron is from the center,
it could be moving around, and there are all kinds of silicon
atoms and bonds and whatnot.
So there are all kinds of stuff between the electron and
the center here.
So I have to use a modified form.
I don't use the permittivity of vacuum.
I map the permittivity of vacuum into
the dielectric constant.
This is the dielectric constant of silicon.
Remember there's so little phosphorus there that it
doesn't alter the properties.
And then for reasons I can't go in to here, I have to
change the mass of the electron into
the effective mass--
we saw this when Bohr did the calculation about the lines of
helium plus--
effective mass of electron in the conduction band.
So if you put both of those in--
and I've got numbers for this--
this value is 11.7 and the second one is about 1/5 of the
rest mass of the electron.
So if you put all of that in, you end up with the energy of
this ground state electron equal to minus k times 0.2
divided by 11.7 squared, which equals minus
0.02 electron volts.
0.02 electron volts.
So this is 2/100 of an electron volt below something.
What's it below?
It's below the conduction band.
It's right here.
This is the ground state of the electron from phosphorus.
It's the ground state of the electron from phosphorus.
E1 of electron from phosphorus.
And phosphorus, the electrical engineers call a donor.
A donor of what?
It's a donor of electrons.
So this thing here is called the donor level.
And that's the ground state.
So the electron should be sitting.
Isn't the ground state here lower than the bottom of the
conduction band?
Sure it is.
So the electron really should be down in here.
There's the electron.
Now you're saying, how are you going to get conductivity?
Well, we just figured out that the energy level of the ground
state of the donor--
so I'm going to call this the energy of the donor level,
ground state--
is 0.02.
How does that compare with thermal energy at room
It's comparable.
So at room temperature with 1/40 of an electron volt,
there's enough energy to promote this electron up into
the conduction band by thermal excitation
from the donor level.
Thermal excitation from the donor level.
And in fact, if we want to do something really, really--
I want to use an adverb here-- really, really cool.
If that's the ground state--
I'm going to blow this up.
So this is the bottom of the conduction band.
And now this is the ground state.
This is the donor level.
This is the donor level, which I was calling E1.

Is it just the donor level here and the conduction band,
or can you see that there's an E2 and an E3 and an E4?
And I have a whole set of quantum states from the donor
level right up to the conduction band.
And if I have enough energy--
if E thermal is on the order of E donor--
I get promotion of these up in here.
Now what would we say if this were the Bohr model?
You'd have an electron loose in a ground state up in here
where it's free to move.
What's that process called?
And you know what the electrical engineers call
these electrons?
They call them ionized electrons.
But they're referring to the ionization of the electron out
of the donor level into the conduction band, not out of
the silicon crystal.
So these are ionized electrons out of donor level.
So we've come full circle.
By the way, this is a calculation.
You know what the measured value is?
It was measured as minus 0.045 electron volts.
You might look at that and say, whoa, that's about two
times this.
Hey, wait a minute.
We started at 13.6.
We went from 13.6 to this, and the real value is this, and so
we are in good shape.
Now, what happens if I put another phosphorus in here?
Another phosphorus, it's also going to sit at this--
I'm going to do it over here.
If I put another phosphorus, it also has
the same donor level.
And another phosphorus, it has the same donor level.
How can I justify this?
Am I not violating the Pauli exclusion principle?
How come all the phosphorus donor states
are at the same level?
Or put another way, under what circumstances is
this diagram accurate?
If the number of phosphorus atoms that I
introduce is tiny--
parts per million--
the separation, the physical separation of this phosphorus
from its nearest phosphorus neighbor is so great, that for
all intents and purposes, they are at infinite separation.
And therefore, all of the phosphorus donor atoms
establish a donor level at the same value.
And if you dope to very, very high levels, you see this
break down.
If you go to very, very high concentrations of phosphorus
dopant, you will discover what happens if these two energy
levels get close enough together in real space.
They have to split.
And so you'll see all of that happening down here at the
miniature level.
The last thing I wanted to do was to push the Bohr model
just a little bit harder.
And so I've given you an energy, and the energy seems
to make sense.
Even though you might say, this is the crazy.
It's not a one-electron system, but it gave us a
reasonable number.
Let's just for interest's sake determine: what's this radius?
I wonder how far this electron is away from the donor
So I plug into the r1.
That's the radius of the ground state electron.
And that's the Bohr radius, right?
0.529 angstroms in elemental hydrogen, but now this has to
be modified by adding the dielectric constant and the
effective mass.
And the number is 30 angstroms, or 3 nanometers, if
you must. All right?
And compare that to silicon-silicon bond distance.
The silicon-silicon bond distance is 2.35 angstroms. So
clearly this electron is very far removed from the central
It's not as I depicted.
It has to be way over here, going way, way around.

This is really, really something.
So now you understand how you can get extrinsic properties
and the behavior of the system governed by the introduction
of impurities.
What I'm going to do next day, is to show you that at typical
doping levels, when you dope, dope at about 10
to the minus 6--
1 part per million--
10 to the minus 6 phosphorus per silicon.
It turns out that the number of electrons from this
operation is much greater than the number of electrons from
thermal excitation, which is tiny.
And therefore, we argue that the properties are governed by
the dopant atom, and so we have extrinsic
behavior that we see.
OK, well that's a day's work.
David, may we go to the slides, please?

OK, so you saw all that last day, and this, this, this.
All right, there's the insulator, and that.
So I'm just doing the lecture again and fast. You can read.
You can parse visual images.
There's the n-type semiconductor.
We're going to look at p-type next day, quickly.
There's a silicon crystal.
There's the salamis--
there's the little wafers that are cut.
And there's the spectrum, the x-ray spectrum indicating you
have a single crystal.
There's the first transistor--
this was a single crystal of germanium
lying here on its side--
the point transistor in the fall of 1947.
Three gentleman at the Bell Labs that demonstrated the
And they won the Nobel Prize for this in 1956.
Now, you can also make
semiconductors that have compounds.
So these are compound semiconductors--
Professor Ballinger showed you last day-- tin, germanium,
silicon, carbon.
And look at all of these.
This is a Group 3 element with a Group 5 element.
5 plus 3 is 8.
Group 3 and 5.
This is Group 2 and 6.
2 plus 6 is 8.
It's always about octet stability.
But they make strong covalent bonds that give band gaps all
over the map.
So suppose I wanted to have something like a stoplight.
Suppose I wanted a stoplight, and I wanted
to make a red light.
Well, I need--
I can go backwards from 660 nanometers, and I can go e
equals hc over lambda and calculate--
I need a material with 1.97 electron volts band gap, and I
look on here and there's nothing that's exactly 1.97.
Well, I can mix these things.
I can mix them.
I can mix something that has 1.52 with 2.3 and get 1.97.
That's called band gap engineering.
And there's more than one mix here, so why would I choose
one mix over another?
It's cheaper, or it's easier to process, or it's stable in
the atmosphere.
You don't want a semiconductor that can't
stand rain and snow.
And so I'm showing you here a variety of compound
semiconductors, and this is the basis of things like--
the LED, the CD reader, the DVD reader is based on light
that's generated from these.
What was the big deal about Blu-ray?
Why is Blu-ray so cool?
Because the original reader was red.
And what's the wavelength of red light?
Red light is around 600 nanometers.
And what's the wavelength of blue light?
It's down around 300 nanometers.
So all other things being equal, my stylus is half the
size, which means for a given area, I can get double the
density of information without changing the size of the disc.
But it wasn't trivial to find a blue laser.
And the person who found the right mix of compounds that
could give blue in an efficient manner became a very
wealthy individual.

So this is our stoplight now.
These are the materials that go into the stoplights.
No more do you have that giant lens with a 150-watt
incandescent bulb.
Now you have the array of these LEDs that are designed
to be in this band gap.
So I think at this point, we've
probably run out of time.
So we'll resume the discussion tomorrow.
Same time, same place.